A gummie bear was tested through a flame-calorimeter test. the bear had a mass of 1.850 grams and the temperature of 100.0 milliliters of water increased by 15.0 degrees celsius. how many calories were in the gummie bear? show all of your calculations.

Answer:

1) 0.18106 M is the molarity of the resulting solution.

2) 0.823 Molar is the molarity of the solution.

Explanation:

1) Volume of stock solution =

Concentration of stock solution =

Volume of stock solution after dilution =

Concentration of stock solution after dilution =

( dilution )

0.18106 M is the molarity of the resulting solution.

2)

Molarity of the solution is the moles of compound in 1 Liter solutions.

Mass of potassium permanganate = 13.0 g

Molar mass of potassium permangante = 158 g/mol

Volume of the solution = 100.00 mL = 0.100  L ( 1 mL=0.001 L)

0.823 Molar is the molarity of the solution.

Final answer:

To determine the molarity of the resulting solution, we can use the formula M1V1 = M2V2. Plugging in the given values, we find that the molarity of the resulting solution is 0.180 MM.

Explanation:

To determine the molarity of the resulting solution, we need to use the formula:

M1V1 = M2V2

Where M1 is the molarity of the stock solution, V1 is the volume of the stock solution used, M2 is the molarity of the resulting solution, and V2 is the final volume of the resulting solution.

Using the given values, we have:

M1 = 0.823 MM

V1 = 11.00 mL

V2 = 50.00 mL

Substituting these values into the formula, we can find the molarity of the resulting solution.

M2 = (M1 * V1) / V2

Plugging in the values:

M2 = (0.823 MM * 11.00 mL) / 50.00 mL = 0.180 MM

The molarity of the resulting solution is 0.180 MM.

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Answer: I believe the answer is an earthquake.

Explanation: Sorry If I am wrong!

Answer:

earthquake

Explanation:

I took the k12 test

Answer:

6.56x10⁻³ mol

Explanation:

First we calculate how many HCl moles are there in 50.0 mL of a 0.1800 M solution:

• 0.1800 M * 50.0 mL = 9.00 mmol HCl

Then we need to calculate how many HCl moles were in excess, that number is the same as the number of NaOH moles they reacted with:

• 0.0980 M * 24.90 mL = 2.44 mmol NaOH = 2.44 mmol HCl

Finally we calculate the difference between the original number of HCl moles and the number remaining after the reaction with XnCO₃:

• 9.00 mmol - 2.44 mmol = 6.56 mmol

• 6.56 mmol / 1000 = 6.56x10⁻³ mol

Final answer:

The amount of HCl that reacted with the carbonate is found by subtracting the amount of HCl neutralized by NaOH from the total initial amount of HCl, resulting in 0.00656 moles of HCl reacting with carbonate.

Explanation:

In this problem, we first need to calculate the total amount of HCl initially used. This is done using the formulaMolarity (M) = moles of solute/Liters of solution. We can calculate moles of HCl as (0.1800 moles/L) * (0.0500 L) = 0.00900 moles.

Next, the HCl reacts with the carbonate, but there is excess that needed to be neutralized using NaOH. The amount of HCl neutralized by NaOH can be calculated similarly as (0.0980 moles/L) * (0.0249 L) = 0.00244 moles.

Thus, the amount of HCl that reacted with the carbonate is the initial moles of HCl minus the moles of HCl neutralized by NaOH. This yields 0.00900 moles - 0.00244 moles = 0.00656 moles of HCl reacted with the carbonate.

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0.0340 g O2

Step 1. Write the balanced chemical equation

4Fe(OH)^(+) + 4OH^(-) + O2 + 2H2O → 4Fe(OH)3

Step 2. Calculate the moles of Fe^(2+)

Moles of Fe^(2+) = 50.0 mL Fe^(2+) × [0.0850 mmol Fe^(2+)/1 mL Fe^(2+)]

= 4.250 mmol Fe^(2+)

Step 3. Calculate the moles of O2

Moles of O2 = 4.250 mmol Fe^(2+) × [1 mmol O2/4 mmol Fe^(2+)]

= 1.062 mmol O2

Step 4. Calculate the mass of O2

Mass of O2 = 1.062 mmol O2 × (32.00 mg O2/1 mmol O2) = 34.0 mg O2

= 0.0340 g O2

0.0342 grams of O2 are consumed to precipitate all of the iron in 50.0 mL of 0.0850 M Fe(II) solution.

To solve this problem, we need to first calculate the number of moles of Fe(II) in 50.0 mL of 0.0850 M Fe(II) solution.

Moles of Fe(II) = (0.0850 mol/L) * (50.0 mL) = 0.00425 mol

According to the balanced chemical equation, 4 moles of Fe(II) react with 1 mole of O2. Therefore, the number of moles of O2 required to precipitate all of the iron in 50.0 mL of 0.0850 M Fe(II) solution is:

Moles of O2 = (0.00425 mol Fe(II)) * (1 mol O2 / 4 mol Fe(II)) = 0.00106 mol O2

Now we can convert the moles of O2 to grams using the molar mass of O2 (32.00 g/mol):

Grams of O2 = (0.00106 mol O2) * (32.00 g/mol) = 0.0342 g O2

Therefore, 0.0342 grams of O2 are consumed to precipitate all of the iron in 50.0 mL of 0.0850 M Fe(II) solution.

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The second option 1,870.4 Joules

Answer:

Four moles of electrons

Explanation:

The reactions in a breathalyzer are redox reactions. Fuel cell breathalyzers consists of fuel cells with platinum electrodes. The current produced depends on the amount of alcohol in the breath. Detection of alcohol involves the oxidation of ethanol to ethanoic acid. The two half cells set in the process are;

Anode;

C2H5OH(aq) + 4OH^-(aq) ----------> CH3COOOH(aq) + 3H2O(l) + 4e

Cathode;

O2(g) + 2H2O(l) +4e--------> 4OH^-(aq)

Hence four electrons are transferred in the process.